MSE 230 Solutions for Assignment 11 Fall 2013 Problem 1 Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration Problem 2 a Hardness testing or tensile testing The yield strength and hardness of 5052 is typically greater than 6061 However this is unreliable because you do not know if the material has undergone the standard strain hardening or heat treating condition If nonstandard conditions were used 5052 may have a lower yield stress than 6061 b X ray diffraction 5052 is a single phase alloy and will have one set of x ray diffraction peaks 6061 is a two phase alloy and one set of peaks will correspond to the lattice parameter associated with the matrix while a second set of peaks will correspond to the second phase particles Problem 3 This problem asks that we specify and compare the microstructures and mechanical properties in the heat affected weld zones for 1080 and 4340 alloys assuming that the average cooling rate is 10 C s Figure 11 28 shows the continuous cooling transformation diagram for an iron carbon alloy of eutectoid composition 1080 and in addition cooling curves that delineate changes in microstructure For a cooling rate of 10 C s which is less than 35 C s the resulting microstructure will be totally pearlite probably a reasonably fine pearlite Figure 11 29 shows the CCT diagram for 4340 steel From this diagram note that a cooling rate of 10 C s produces a totally martensitic structure Pearlite is softer and more ductile than martensite and therefore is most likely more desirable in the case of a weld Problem 4 Callister 11 D14 This problem is concerned with the precipitation hardening of copper rich CuBe alloys It is necessary for us to use the Cu Be phase diagram Figure 11 30 a The range of compositions over which these alloys may be precipitation hardened is between approximately 0 2 wt Be the maximum solubility of Be in Cu at about 300 C and 2 7 wt Be the maximum solubility of Be in Cu at 866 C b The heat treatment procedure of course will depend on the composition chosen First of all the solution heat treatment must be carried out at a temperature within the phase region after which the specimen is quenched to room temperature Finally the precipitation heat treatment is conducted at a temperature within the 2 phase region For example for a 1 5 wt Be 98 5 wt Cu alloy the solution heat treating temperature must be between about 600 C and 900 C while the precipitation heat treatment 1 would be below 600 C and probably above 300 C Below 300 C diffusion rates are low and heat treatment times would be relatively long Problem 5 From the class lecture notes 6 E T 70 000MPa 3 10 C 100 C c 26 3 MPa 1 1 0 2 Recall in class we assumed that glass had a strength c of 70 MPa The actual strength of glass is greatly affected by surface flaws i e scratching glass lowers the strength making it easy to cut glass Thus the apparent decrease in strength 70 MPa to 26 3 MPa with exuberant scrubbing was most likely due to scratches introduced during scrubbing Critical flaw size K a 2 1 K 2 1 0 7 MPa m 4 a c IC 2 25x10 m 26 3 MPa or 225 m to prevent this decrease in strength physically polish away the larger surface flaws chemically etch the glass to smooth out the scratches fire polishing heat the glass causing localized flow to smooth out the scratches Problem 6 There are a couple of ways to approach this problem Approach 1 As the interior layer cools it wants to shrink but is prevented from doing so by the already rigid outer layer The amount of thermal contraction that the interior layer would undergo if unconstrained is given by T For pyrex 3x10 6 C 680 C 2 04 x 10 3 Since the interior is constrained the residual tensile stress is E 2 04 x 10 3 70 x 103 MPa 143 MPa This residual tensile stress on the interior causes a larger surface compressive stress since the 2 surface layer is thinner than the interior layer The surface compressive stress can be estimated by thickness of exterior layer ext thickness of interior layer int ext 6 mm 143 MPa 4 mm ext 215 MPa Repeating this calculation for soda lime glass gives int 476 MPa tensile stress ext 714 MPa compressive stress much larger due to the larger coefficient of thermal expansion Approach 2 You should come up with a similar estimate in fact somewhat larger applying the equation for thermal stresses from Prob 2 to this situation Temperature Stress Distribution 20 C 2 8 476 MPa Stress MPa Temperature C 700 C Distribution 143 MPa Position mm 10 215 MPa 10 714 MPa Position mm Pyrex Soda Lime 3 Glass Problem 7 a 60 of theoretical density volume fraction of porosity 0 40 From Chap 12 11 equation 12 9 E Eo 1 1 9P 0 9P2 393 GPa 1 1 9 0 4 0 9 0 4 2 E 151 GPa Modulus of Elasticity GPa b When the powder compact is placed in the furnace at 0 95Tm it will undergo densification via sintering see Chap 13 11 pages 524 525 Therefore the volume fraction of porosity will decrease and the modulus of the powder compact will increase see Fig 12 35 on page 489 of Callister The plot of modulus of elasticity vs time should be asymptotic to E0 because the modulus of elasticity of the powder compact will not exceed the Young s modulus of Al2O3 E0 Time Problem 8 Refer to the SiO2 Al2O3 phase diagram Fig 12 27 on page 480 of Callister a 70 wt Al2O3 30 wt SiO2 Minimum temperature to form liquid 1588 C Weight fraction of liquid First let s get the phases and phase compositions from the mullite and liquid two phase region Mullite 71 wt Al2O3 Liquid 8 wt Al2O3 4 WL C MUL CO 71 70 0 016 C MUL C L 71 8 or 1 6 wt liquid sounds good for liquid phase sintering b 80 wt Al2O3 20 wt SiO2 Minimum temperature to form liquid 1891 C Phases and compositions Alumina Al2O3 100 wt Al2O3 Liquid 75 wt Al2O3 WL C Al2 O3 C O C Al 2 O3 C L 100 80 0 80 100 75 or 80 wt liquid sounds lousy for liquid phase sintering because most of the material present is liquid and very little is solid Problem 9 a The specific surface area decreases rapidly with increasing particle size Note that the log scale makes it much easier to follow the trend This plot suggests that smaller particles have a much greater driving force for sintering because reducing surface energy area is the driving force for sintering and the smaller particles have a lot more surface area to lose in contrast to the larger particles 5 …
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